@Brilliant Member
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Ya I got kv^2/2.it will be misleading for many I guess to mark kv^2.but if the part held goes a distance of vt then the part coming to motion will have a length of vt/2.its quite apparent and easy to analyse if the part held goes vt then the part that will be available to come to motion must be vt as length is conserved but this appears in the curl with 2 half length of (vt/2)each!!!so now we can use f=vdm/dt =vd(kvt/2)/dt=kv^2/2.

shit ! i wrote it on paper and clicked pic to show you , displaced the power cord a little and boom , computer off. after then solved 4 level 5's and wrote their solution , upvote them too :P le dekh le